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For nitrous acid (hno2), the value of ka is 4.5 x 10-4. rn a buffer solution was prepared by dissolving 0.085 moles ofrn nitrous acid and 0.100 moles of potassium nitrite (kno2) rn in enough water to make 1.00 liter of solution. rn What is the ph of this buffer solution?rn Henderson-Hasselbalch Equation: ph = pka + log ( [a-1] / [ha] )
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SOLUTION
The Henderson Hasselbalch equation is written as:
$$pH = pKa + log frac{[A^-]}{[HA]}$$
where
* $pKa$ is the negative logarithm of the acid dissociation constant ($Ka$)
* $[A^-]$ is the molar concentration of the conjugate base
* $[HA]$ is the molar concentration of the weak acid
For this problem, we have the following information:
* $pKa = -log(Ka) = -log(4.5 times 10^{-4}) = 3.35$
* $[A^-] = 0.100 M$ (concentration of nitrite ion, $NO_2^-$)
* $[HA] = 0.085 M$ (concentration of nitrous acid, $HNO_2$)
Plugging these values into the Henderson Hasselbalch equation, we get:
$$pH = 3.35 + log frac{0.100 M}{0.085 M} = 3.46$$
Therefore, the pH of the buffer solution is 3.46.
The Henderson Hasselbalch equation is written as:
$$pH = pKa + log frac{[A^-]}{[HA]}$$
where
* $pKa$ is the negative logarithm of the acid dissociation constant ($Ka$)
* $[A^-]$ is the molar concentration of the conjugate base
* $[HA]$ is the molar concentration of the weak acid
For this problem, we have the following information:
* $pKa = -log(Ka) = -log(4.5 times 10^{-4}) = 3.35$
* $[A^-] = 0.100 M$ (concentration of nitrite ion, $NO_2^-$)
* $[HA] = 0.085 M$ (concentration of nitrous acid, $HNO_2$)
Plugging these values into the Henderson Hasselbalch equation, we get:
$$pH = 3.35 + log frac{0.100 M}{0.085 M} = 3.46$$
Therefore, the pH of the buffer solution is 3.46.
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